C# 这些 ( += , -= , *= , /= ) 运算符是什么意思？
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What do these ( += , -= , *= , /= ) Operators Mean?
提问by Ian Lundberg
I have been looking everywhere to figure out what these mean and how they are used +=, -=, *=, /=, the most I have found is that they are, "Assignment by Addition", "Assignment by Difference", "Assignment by Product", "Assignment by Quotient", etc, but I can't figure out when or how they are used. If anyone can please explain this to me I would be very grateful. thanks
我一直在到处找弄清楚这些意味着什么，他们是如何使用的+=，-=，*=，/=，我已经找到了最重要的是，他们是“分配添加”，“差量分配”，“按产品转让”，“转让按商数”等，但我无法弄清楚它们何时或如何使用。如果有人可以向我解释这一点，我将不胜感激。谢谢
采纳答案by Totero
They are shorthand:
它们是速记：
a += b
a += b
is the same as
是相同的
a = a + b
a = a + b
Etc...
等等...
so
所以
a -= bis equivalent toa = a - ba *= bis equivalent toa = a * ba /= bis equivalent toa = a / b
a -= b相当于a = a - ba *= b相当于a = a * ba /= b相当于a = a / b
As Kevin Brydon suggested - Familiarize yourself with the operators in C# here.
正如 Kevin Brydon 建议的那样 - 在此处熟悉 C# 中的运算符。
回答by Ravindra Bagale
these are shorthand operators.
these are used when you does the operation & stores result into one of the variable between them. that is you store result into one of your operand
suppose example
1)x=x+y;
here you can do x+=y;
ex 2) x=x+1;
here you can do x+=1;
这些是速记运算符。
当您执行操作并将结果存储到它们之间的变量之一时会使用这些。也就是说，您将结果存储到您的操作数之一中，假设示例
1)x=x+y;
在这里你可以做 x+=y;
例 2) x=x+1;
在这里你可以做 x+=1;
回答by Michael Krelin - hacker
Roughly, var *operator*= expressionmeans var = var *operator* expression. Also, I've heard there's a documentation somewhere.
粗略地说，var *operator*= expression意味着var = var *operator* expression。另外，我听说某处有文档。
回答by andy
回答by wxyz
a+=1 means a = a+1
a-=2 means a = a-2
a*=3 means a = a*3
a/=4 means a = a/4
回答by Rawling
See 7.13 Assignment operatorsin the specand its subsections., specifically 7.13.2 Compound assignment:
请参阅7.13 Assignment operators规范及其小节，特别是7.13.2 Compound assignment：
An operation of the form x op= y is processed by applying binary operator overload resolution (Section 7.2.4) as if the operation was written x opy. Then,
?If the return type of the selected operator is implicitly convertible to the type of x, the operation is evaluated as x = x opy, except that x is evaluated only once.
?Otherwise, if the selected operator is a predefined operator, if the return type of the selected operator is explicitly convertible to the type of x, and if y is implicitly convertible to the type of x, then the operation is evaluated as x = (T)(x opy), where T is the type of x, except that x is evaluated only once.
?Otherwise, the compound assignment is invalid, and a compile-time error occurs.
x op= y形式的运算是通过应用二元运算符重载解析（第 7.2.4 节）来处理的，就好像该运算被写成 x opy 一样。然后，
? 如果所选运算符的返回类型可隐式转换为 x 的类型，则该操作的计算结果为 x = x opy，但 x 仅计算一次。
?否则，如果所选运算符是预定义的运算符，如果所选运算符的返回类型可显式转换为 x 的类型，并且如果 y 可隐式转换为 x 的类型，则该操作的计算结果为 x = ( T)(x opy)，其中 T 是 x 的类型，只是 x 只计算一次。
?否则复合赋值无效，会出现编译时错误。
回答by Mike Parkhill
It's a short form. So instead of writing:
这是一个简短的形式。所以而不是写：
x = x + 1;
x = x + 1;
You can simply write:
你可以简单地写：
x += 1;
x += 1;
It has the same affect.
它具有相同的影响。
